Consider a set of distinct numbers arranged in some
geometrical pattern on the plane such that every sum of co-linear numbers
through the pattern has the same value. The most common example is a magic
square, which consists of a square arrangement of numbers such as the 3-by-3
arrangement shown below.

It’s customary to require the sums along each of the main
diagonals to also equal the common sum, but even without those diagonals
conditions we already get interesting structure from just the six orthogonal
sums, especially if we impose some numerical constraints on the individual
numbers. For example, we might seek a square arrangement of perfect square
integers. The smallest such example is

where the common sum is S = 3249 = 572. As
discussed in the note on Orthomagic Squares of
Squares, this is just one of an infinite family of such arrangements. (On
the other hand, if we impose the diagonal conditions as well, there are no
known 3x3 magic squares of squares.)

It’s interesting to consider other planar arrangements of
nine numbers with six specified sums required to have the same value. One
such pattern is the nested triangular arrangement – which we call an Amanda arrangement
– as shown below.

In this pattern each of the six indicated sums of
co-linear terms is 12. Several interesting symmetries appear in this
arrangement. Note, for example, that each of the pairs (1,7), (2,8), and
(3,9) differ by 6, and if we transpose each of these pairs the net effect is
to increase each of the six indicated sums by 6. (Each outer sum is
incremented by +6, and each inner sum is incremented by +6-6+6.) Hence we have the “dual” arrangement

To show that every Amanda Arrangement has such a dual, let
us assign symbols to the terms as shown in the figure below.

We have the following six equations in the nine variables

Each of these sets of three equations can be solved for
the B terms as follows.

Subtracting A1, A2, and A3
respectively from both sides of the three equations on the left, and
subtracting C1, C2, and C3 respectively from
both sides of the three equations on the right, these equations can be
written as

for j = 1, 2, 3, where for convenience we have defined the
parameters

Thus the three differences Bj – Aj
are all equal, as are the three differences Bj – Cj.
The latter equalities are sufficient to ensure that each Amanda Arrangement
has a dual of the kind described previously, i.e., transposing Bj
and Cj. Furthermore, summing each of equations (1) over the index
j gives

Subtracting S-a
from both sides of the left hand equation, and subtracting S-g from both sides of the right equation,
we get

Substituting back into equation (1) gives

and hence

In our original example, each of the opposite pairs of the
inner triangle differed by 6, and each of the opposite pairs of the outer
triangle differed by 3. The above equation shows that this ratio of 2 applies
to all Amanda arrangements. We also note that this equation is symmetrical in
Bj and Cj, since it can be written in the form

Now suppose we require the individual numbers to be
perfect squares. It isn’t obvious that any such arrangements exist, but
in fact they do, although they are fairly rare. One way of seeking such
arrangements is to examine the positive integers in sequence, and for each
integer N determine the two-part factorizations N = xy where x and y are
integers with the same parity and with x greater than y. Then for each such
factorization compute b = (x+y)/2 and c = (x-y)/2,
so we have N = b2 – c2. If an integer N has three
distinct factorizations such that a2 = (b2 + c2)/2
is a square integer, then these numbers represent an Amanda arrangement of
squares, putting Aj = aj2, Bj = bj2,
and Cj = cj2.

For example, the integer N = 6720 has (among others) the
three factorizations

The first of these factorizations gives b = (224+30)/2 =
127 and c = (224-30)/2 = 97, and we
find that 1272 + 972 = 2(1132), so a = 113.
Likewise the other two factorizations listed above give integer values of a,
so we have found the smallest Amanda arrangement of squares shown below.

The common sum for this arrangement is S = 18249, whereas
for the dual arrangement (found by transposing the inner opposite pairs) is
S’ = 24969. The difference between these two common sums is 6720 = (26)(3)(5)(7),
which as we’ve seen is also the common difference between the opposite
numbers for the inner triangle. Of course, multiplying all the base numbers
by a constant trivially gives another Amanda arrangement of squares, as does
rotating the numbers by 120 degrees or reflecting about a suitable axis, but
the results are not essentially different from this one or its dual.

The next smallest primitive Amanda arrangement of squares is
based on the factors of N = 87350, leading to the arrangement shown below.

For this arrangement the common sum is S = 213369, and the
common sum of the dual arrangement is S’ = 300729. As always, the difference
between these two sums, which is the common difference between the opposite
elements of the inner triangle, i.e., 87360.

Examining all the integers N less than ten million, we
find that there are only nine “Amanda numbers”, i.e., numbers with three (or
more) acceptable factorizations leading to Amanda arrangements. One of these
numbers (N = 27350040) actually has four acceptable factorizations, so it
serves as the basis of four distinct primitive Amanda arrangements. The
fourteen Amanda numbers less than twenty million, along with their prime
factorizations, are listed below.

Each of these is divisible by 26. The
arrangements corresponding to the first two of these Amanda numbers have
already been depicted. The square roots of the vertex values for the
arrangements corresponding to the next six values, excluding 2735040, are as
follows:

The squares of the elements in each row of any of these
3x3 arrays are in arithmetic progressions (since ci2 +
bi2 = 2ai2), and the differences
between the squares of the values in neighboring columns are the same for
each row. This is reminiscent of the conditions on a 3x3 array of squares
that would be necessary for a magic square of
squares, since we can arrange the numbers in a 3x3 magic square so that
each row and each column is an arithmetic progression. However, there are no
(known) 3x3 magic squares of squares, whereas we have found these several
examples of Amanda arrangements.

In the above tabulations we left out the case N = 2735040
because, as noted above, for that case there are four (rather than just
three) acceptable factorizations. Hence we could construct four distinct
Amanda arrangements from this number. However, it’s more natural to regard
this as a three-dimensional tetrahedral arrangement, as depicted below, where
the sum of the vertices on each of the indicated planes has the same value.

The values of the vertex bases for this tetrahedral
arrangement are listed below.

The sum of the four vertices in each of the eight planes
comprising the faces of the two outer tetrahedra equals S = 56192890. For
example, we have

This is the only tetrahedral Amanda arrangement of squares
with N less than twenty million. Presumably there are infinitely many such
arrangements, but they are evidently fairly rare. Only one other number with four
acceptable factorizations is known (see below), and no number with five or
more acceptable factorizations is known.

Above 17 million there seems to be a large gap, and the
next two smallest primitive Amanda numbers are over 50 million:

We also find a number that is a square multiple of a
previous Amanda number, so it isn’t primitive, but the previous number has
only three acceptable factorizations, whereas if we multiply it by 4 we get

which has four acceptable factorizations, leading to the
tetrahedral vertices shown below.

The first three of these triples are just multiples of the
original triples, but the fourth is new, so this gives a second tetrahedral
Amanda arrangement.

Another way in which we might seek to extend these
arrangements is by adding more triangular levels, as indicated in the figure
below.

Any such arrangement would require the relations

However, it isn’t self-evident that there is even one
solution of this pair of equations, let alone three. This is closely related
to the problem of finding four squares in arithmetic progression, which is
impossible, as shown in another note.
But equations (3) imply a slightly different condition, namely, four squares
that occupy four of five consecutive terms in arithmetic progression.
Specifically, if we require each of the vertices to be a square integer,
equations (3) for each index require integers a,b,c,d such that

It follows that c2 is half-way between a2
and d2, and b2 is half way between a2 and c2,
as depicted below.

m

The smallest set of four such values is

If we examine the other two possible ways of distributing
four numbers over five consecutive terms of an arithmetic progression, it can be shown that no solutions exist for
the case when the unoccupied term is in the middle, but we do find a solution
for the case when the unoccupied term is at the low end, as depicted below.

This corresponds to the equations

The smallest primitive solution of this pair of equations
is

This solution is closely related to the previous one, as
discussed in another note, where we
discuss how to generate an infinite family of primitive solutions to (4) and
(5). To explore some other aspects of the relationship between these two
solutions, note that the expressions for the differences of the first
solution in terms of the values of the second solution are as shown below.

Also, we can factor both sides of the relation c2
– d2 = 2(b2 – c2) to give

Substituting for c-d
and b-c and dividing through by a, we get

Combining this with the equations for a-b, b-c,
and c-d, we get the four equations in
four unknowns

Solving this system of equations gives

where

Inserting the values of the parameters for the only known
solution (23,17,13,7), this gives Q = 5730. Of course, the four conditions
leading to this result are not algebraic identities, they are just numerical coincidences,
so (6) is not an algebraic solution of (4). It happens to be a solution if we
set a, b,
g, d
to the values for the one known solution of (5). However, even if a, b,
g, d
satisfy (5), they still do not make (6) an algebraic solution. To identify
the further condition that must be satisfied by a, b, g, d
in order to make (6) a solution, we consider another approach to the above
expression for Q, which is to simply postulate a solution of the form (6) and
then substitute those expressions into (4) to give the following two
equations for Q:

The algebraic solutions of these two equations,
respectively, are

These two expressions for Q are not generally equal to
each other for arbitrary choices of the parameters. On the other hand, if we
solve for the single value of Q that makes the left-hand sides of (7) equal
to each other, we get the expression for Q beneath equation (6), but this
does not generally give solutions of the individual equations (7).
Interestingly, if we set the sum (instead of the difference) between the left
sides of (7) to zero (a condition that would also be satisfied by genuine
solution of those equations), and solve for Q, we get the more symmetrical
expression

If substitute this expression for Q into the left side of
the first of equations (7) we find (not surprisingly) that it does not vanish
identically, but if we replace each occurrence of a2 with the equivalent 2b2 – d2,
and then replace each b2
with the equivalent 2g2 – d2, we find that the resulting
expression factors as

Thus we have a solution if and only if (a-d) = 4(b-g).
Noting that equations (5) imply

it follows that an equivalent condition is

This happens to be satisfied by the one known solution of
(5), namely, a = 23, b = 17, g
= 13, d = 7, so this explains why we
can generate a solution of (4) based on this solution of (5). In general, any
solution of (5) that also satisfies (8) can be used to generate a solution of
(4). Substituting for d in equations
(5) and adding the results, we find that b
must equal 3a – 4g, and if we make that substitution both
equations reduce to

Multiplying through by 13 and completing the square, this
can be written as

and hence 13a = 23g. This shows that there are no other dual
solution sets of equations (4) and (5) of the type described here.

There is, however, another interesting structural aspect
of these relations. In the preceding discussion we made use of the roots of
the difference between the squares of the left sides of equations (7),
because we considered the sum and the difference of those quantities. It
might seem that we could force both quantities to equal zero by setting the sum
of their squares to zero. Solving the resulting quadratic in Q leads to the rational
expression

This could be regarded as a an rational algebraic
pseudo-solution, because for any arbitrary values of the parameters it makes
the sum of the squares of the left sides of equations (7) vanish. The
parameters need not even satisfy (5). If we were concerned only with
real-valued numbers, this would indeed imply a complete solution, giving an
infinite four-parameter rational algebraic solution of the pair of
Diophantine equations (4). However, notice that this expression for Q is
generally complex, so the vanishing of the sum of the squares of two
quantities does not imply that each quantity individually vanishes. It so
happens that Q has the purely real value 5730 for the parameter values of the
single known solution, but for general parameters values we get a complex Q,
and neither of equations (7) are individually satisfied. A complete solution
requires values of the parameters that make Q real, and a sufficient
condition for this to occur is for the parameters to satisfy (7) with a+d
= b+g
as noted previously.